SICP中sqrt（开方）的实现（附C#实现）

#lang racket
(define (square x) (* x x))
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
(define (improve guess x)
  (average guess (/ x guess)))

(define (average x y)
  (/ (+ x y) 2))

(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.001))

(define (sqrt x)
  (sqrt-iter 1.0 x))


(sqrt 9)

3.000000001396984
> 

此为SICP中给出的实现方法，使用常规的过程调用能力实现。

下为我在C#中的实现方法。

static void Main(string[] args)
{
double num1;
int num2;
Console.WriteLine("请输入要开方的值：");
num1 = float.Parse(Console.ReadLine());
Console.WriteLine("请输入精度");
num2 = Int32.Parse(Console.ReadLine());
num1 = sqrt(num1,num2);
Console.WriteLine(num1);
Console.ReadLine();
}
static double sqrt(double num,int num1)
{
double precision = Math.Pow(0.1, num1);//定义精度
double result = 1;//定义结果
while (Math.Abs( Math.Pow((num / result + result)/2, 2) - num) > precision)//
{

result = (num / result + result) / 2;
}

return result;
}

 SICP中用比例法改进的算法

#lang racket
(define (square x) (* x x))
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
(define (improve guess x)
  (average guess (/ x guess)))

(define (average x y)
  (/ (+ x y) 2))

(define (good-enough? guess x)
  (< (abs (- (/ (square guess) x) 1) ) 0.00001))

(define (sqrt x)
  (sqrt-iter 1.0 x))


(sqrt 0.002)